Answer:
(a) m = 33.3 kg
(b) d = 150 m
(c) vf = 30 m/s
Explanation:
Newton's second law to the block:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Data
F= 100 N
a= 3.0 m/s²
(a) Calculating of the mass of the block:
We replace dta in the formula (1)
F = m*a
100 = m*3
m = 100 / 3
m = 33.3 kg
Kinematic analysis
Because the block moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+a*t Formula (3)
Where:
d:displacement in meters (m)
t : time interval in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
a= 3.0 m/s²
v₀= 0
t = 10 s
(b) Distance the block will travel if the force is applied for 10 s
We replace dta in the formula (2):
d= v₀t+ (1/2)*a*t²
d = 0+ (1/2)*(3)*(10)²
d =150 m
(c) Calculate the speed of the block after the force has been applied for 10 s
We replace dta in the formula (3):
vf= v₀+a*t
vf= 0+(3*(10)
vf= 30 m/s
Answer:
theroy of plate tectonics
A steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.the speed of transverse waves on the wire would be 372.77 m/s
<h3>What is a sound wave?</h3>
It is a particular variety of mechanical waves made up of the disruption brought on by the movements of the energy. In an elastic medium like the air, a sound wave travels through compression and rarefaction.
For calculating the wave velocity of the sound waves generated from the piano can be calculated by the formula
V= √F/μ
where v is the wave velocity of the wave travel on the string
F is the tension in the string of piano
μ is the mass per unit length of the string
As given in question a steel piano wire, of length 1.150 m and mass of 4.80 g is stretched under a tension of 580.0 N.
The μ is the mass per unit length of the string would be
μ = 4.80/(1.150×1000)
μ = 0.0041739 kg/m
By substituting the respective values of the tension on the string and the density(mass per unit length) in the above formula of the wave velocity
V= √F/μ
V=√(580/0.0041739)
V = 372.77 m/s
Thus, the speed of transverse waves on the wire comes out to be 372.77 m/s
Learn more about sound waves from here
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An electric engine turning a workshop sanding rotation at 1.00 × 10² rev/min is switched off. Take the wheel includes a regular negative angular acceleration of volume 2.00 rad/s². 5.25 moments long it takes the grinding rotation to control.
<h3>What is negative angular acceleration?</h3>
- A particle that has a negative angular velocity rotates counterclockwise.
- Negative angular acceleration () is a "push" that is hence counterclockwise.
- The body will speed up or slow down depending on whether and have the same sign (and eventually go in reverse).
- For instance, when an object rotating counterclockwise slows down, acceleration would be negative.
- If a rotating body's angular speed is seen to grow in a clockwise direction and decrease in a counterclockwise direction, it is given a negative sign.
- It is known that a change in the linear acceleration correlates to a change in the linear velocity.
Let t be the time taken to stop.
ω = 0 rad/s
Use the first equation of motion for rotational motion
ω = ωo + α t
0 = 10.5 - 2 x t
t = 5.25 second
To learn more about angular acceleration, refer to:
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Answer:
a) 34.05ft/s
b).1156.2BTU/lbm
c) 2.04BTU/s
Explanation:
Amount of liquid that has evaporated, m = ◇Vliq/ Vf
We replace the values to make conversion
m = (0.6gal/ 0.01683ft^3/lbm) × (0.13368ft^3/1gal)
m = 4.755lb
The mass flow rate of exit steam is given by:
m' = m/◇t
We replace values to make conversion
m' =( 4.766lb/45min) = 0.1059lb/min × 1min/60s
m' = 0.001765lb/s
The exit velocity V = m'/pA = m'Vg/A
We replace values to make conversion
V =[ (0.001765lbm/s)(20.093ft^3/lbm) /(0.15 in^2)]× (144in^2/1ft^2)
V = 34.05ft/s
b) The total and flow energies per unit mass is given by:
Eflow= Pv = h - u
We replace the values to make conversion
Eflow = 1156.2 - 1081.8
Eflow = 74.4BTU/lbm
Therefore theta= h + ke + pe
Theta approximately =h = 1156.2BTU/lbs
c) The rate at which energy is leaving the cooler by steam is given by:
Emass = m'theta
Emass = (0.001765)×(1156.2)
Emass = 2.04BTU/s
